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  1. #1
    Member Em Yeu Anh's Avatar
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    Arrow Finding a power series

    Question: Expand \frac{1}{(1+x)^\frac{1}{4}} as a power series and use it to estimate \frac{1}{1.1^\frac{1}{4}} correct to three decimal places.

    I am clueless!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Em Yeu Anh View Post
    Question: Expand \frac{1}{(1+x)^\frac{1}{4}} as a power series and use it to estimate \frac{1}{1.1^\frac{1}{4}} correct to three decimal places.

    I am clueless!
    Binomial series for non-integer exponents:

    (1+x)^{\frac{1}{4}}= ..

    Then plug in x=0.1, and as the series is alternating the absolute value of the error is less than the absolute value of the first ignored term

    CB
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  3. #3
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    Quote Originally Posted by Em Yeu Anh View Post
    Question: Expand \frac{1}{(1+x)^\frac{1}{4}} as a power series and use it to estimate \frac{1}{1.1^\frac{1}{4}} correct to three decimal places.

    I am clueless!
    That is f(x)= (1+ x)^{-1/4}

    f'(x)= -\frac{1}{4}(1+ x)^{-5/4}

    f"(x)= \frac{5}{16}(1+ x)^{-9/4}

    f'''(x)= -\frac{45}{64}(1+ x)^{-13/4}

    The nth derivative is, I believe, \frac{(-1)^n}{1(3)(9)\cdot\cdot\cdot(4n+1)}(1+ x)^{-(4n+1)/4}.

    Write it as a Taylor's series about x= 1. Since the sign is alternating, the error is never larger than the next term in the series. Expand it until the next term is less than .001.

    (Well, yeah, the generalized binomial series is probably easier!)
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  4. #4
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    Quote Originally Posted by Em Yeu Anh View Post
    Question: Expand \frac{1}{(1+x)^\frac{1}{4}} as a power series and use it to estimate \frac{1}{1.1^\frac{1}{4}} correct to three decimal places.

    I am clueless!
    Also check out "Binomial Theorem" on Wikipedia. Rising factorial, falling factorial, I'm confused. How about I show you this Mathematica code and then you figure out how I got it and what it means.

    Code:
    n[858]:=
    myFunctions[x_] := 1/(1 + x)^(1/4)
    mySeries[x_, n_] := 
      Sum[(FactorialPower[-4^(-1), k]/k!)*
        x^k, {k, 0, n}]
    mySeries[0.1, 20]
    myFunctions[0.1]
    
    Out[860]=
    0.9764540896763106
    
    Out[861]=
    0.9764540896763105
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